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(D) The intensity of radiation emitted from the surface of a star is proportional to its projected area.Let $I \propto A$,where $A$ is the area of the

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the minimum observed intensity of the parent star is $0.9999 \,I_0$

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(D) The intensity of radiation emitted from the surface of a star is proportional to its projected area.Let $I \propto A$,where $A$ is the area of the star's disc.If $I_0$ is the intensity of the parent star,then $I_0 = k \pi (100 R)^2 = k \pi R^2 	imes 10000$,where $k$ is a constant.When the exoplanet is in front of the star,the observed intensity is at its minimum $(I_{\min })$. This occurs because the exoplanet blocks a portion of the star's disc equal to its own cross-sectional area $\pi R^2$.Thus,$I_{\min } = k [\pi (100 R)^2 - \pi R^2]$$I_{\min } = k \pi R^2 (10000 - 1) = k \pi R^2 	imes 9999$Taking the ratio of the minimum intensity to the original intensity:$\frac{I_{\min }}{I_0} = \frac{k \pi R^2 	imes 9999}{k \pi R^2 	imes 10000} = \frac{9999}{10000} = 0.9999$Therefore,$I_{\min } = 0.9999 \,I_0$.

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